Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A tuning fork of \(512 \mathrm{~Hz}\) is used to produce resonance in a resonance tube experiment. The level of water at first resonance is $30.7 \mathrm{~cm}\( and at second resonance is \)63.2 \mathrm{~cm}$. What is the error in calculating velocity of sound? Assume the speed of sound $330 \mathrm{~m} / \mathrm{s}$. (A) \(58(\mathrm{~cm} / \mathrm{s})\) (B) \(204.1(\mathrm{~cm} / \mathrm{s})\) (C) \(280(\mathrm{~cm} / \mathrm{s})\) (D) \(110(\mathrm{~cm} / \mathrm{s})\)

Short Answer

Expert verified
The error in calculating the velocity of sound is \(110(\mathrm{~cm} / \mathrm{s})\).

Step by step solution

01

Calculate the fundamental frequency of the tube

To calculate the fundamental frequency, we need to use the difference in water level at the first and second resonances, which is \( 63.2~cm - 30.7~cm = 32.5~cm \). We also need the frequency of the tuning fork, which is \(512Hz\). The relationship between the frequency of the tuning fork, the fundamental frequency, and the difference in water level is: \( n\Delta L = 2(2L_1 - \Delta L) \) Where \(n\) is the number of resonances (\(n = 2\) for the first resonance, and \(n = 3\) for the second resonance) and \(L_1\) is the water level at the first resonance. Since we only need to calculate the fundamental frequency, we'll use the data for the first resonance. For \(n=2\), the fundamental frequency (\(f_1\)) is given by: \( f_1 = \dfrac{f}{n-1} \) Where \( f \) is the frequency of the tuning fork.
02

Calculate the experimental speed of sound

Now that we have the fundamental frequency, we can calculate the experimental speed of sound using the water level at first resonance. Experimental speed of sound is given by: \(v = 2L_1 f_1 \)
03

Calculate the error in the experimental speed of sound

To find the error in the experimental speed of sound, we need to compare it with the given value, which is \(330 m/s\). The percentage error can be calculated as: \( \% \text{ error } = \dfrac{ \text{ Experimental speed of sound } - \text{ Actual speed of sound}}{ \text{ Actual speed of sound}} \times 100 \)
04

Convert the error to cm/s and find the answer

Now that we have the percentage error, we need to convert it to cm/s: \( \text{ Error } = ( \% \text{ error }\times \text{ Actual speed of sound}) / 100~cm/s \) Compare the error with the given options. The one closest to the calculated error is the answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A closed organ pipe of length \(\mathrm{L}\) and an open organ pipe contain gases of densities \(\rho_{1}\) and \(\rho_{2}\) respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. What is the length of the open organ pipe? (A) \(\left.(4 \mathrm{~L} / 3) \sqrt{(} \rho_{1} / \rho_{2}\right)\) (B) \((\mathrm{L} / 3)\) (C) \((4 \mathrm{~L} / 3) \sqrt{\left(\rho_{2} / \rho_{1}\right)}\) (D) (4L / 3)

An air column in a pipe, which is closed at one end will be in resonance with a vibrating tuning fork of frequency \(264 \mathrm{~Hz}\), What is the length of the column if it is in \(\mathrm{cm} ?\) (speed of sound in \(\operatorname{air}=330 \mathrm{~m} / \mathrm{s}\) ) (A) \(62.50\) (B) \(15.62\) (C) 125 (D) \(93.75\)

A hollow pipe of length \(0.8 \mathrm{~m}\) is closed at one end. At its open end a \(0.5 \mathrm{~m}\) long uniform string is vibrating in its second harmonic and it resonates with the fundamentals frequency of the pipe if the tension in the wire is \(50 \mathrm{~N}\) and the speed of sound is $320 \mathrm{~ms}^{-1}$, what is the mass of the string. (A) \(20 \mathrm{~g}\) (B) \(5 \mathrm{~g}\) (C) \(40 \mathrm{~g}\) (D) \(10 \mathrm{~g}\)

An open pipe is in resonance in \(2^{\text {nd }}\) harmonic with frequency \(\mathrm{f}_{1}\). Now one end of the tube is closed and frequency is increased to \(\mathrm{f}_{2}\) such that the resonance again occurs in nth harmonic. Choose the correct option. (A) \(\mathrm{n}=5, \mathrm{f}_{2}=(5 / 4) \mathrm{f}_{1}\) (B) \(\mathrm{n}=3, \mathrm{f}_{2}=(3 / 4) \mathrm{f}_{1}\) (C) \(\mathrm{n}=5, \mathrm{f}_{2}=(3 / 4) \mathrm{f}_{1}\) (D) \(\mathrm{n}=3, \mathrm{f}_{2}=(5 / 4) \mathrm{f}_{1}\)

A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency \(512 \mathrm{~Hz}\). If the tube is opened at both ends what is the fundamental frequency that can be excited (in \(\mathrm{Hz}\) )? (A) 256 (B) 1024 (C) 128 (D) 512

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free